TLS Online TPP Program

#Question id: 18013


Each of five questions on a multiple-choice examination has four choices, only one of which is correct. The student is attempting to guess the answers. The random variable X is the number of questions answer correctly. Probability that the student will get At least one correct answer.

#Part-A Aptitude & General Biotechnology
  1. 81/128
  2. 781/1024
  3. 63/64
  4. 57/224
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TLS Online TPP Program

#Question id: 14160

#Part-B Specialized Branches in Biotechnology

In FASTA, For a Z-score > 15, the match can be considered extremely ______ with _____ of a homologous relationship.

TLS Online TPP Program

#Question id: 14160

#Part-A Aptitude & General Biotechnology

In FASTA, For a Z-score > 15, the match can be considered extremely ______ with _____ of a homologous relationship.

TLS Online TPP Program

#Question id: 14161

#Part-B Specialized Branches in Biotechnology

Which of the following is not a description of dynamic programming algorithm?

TLS Online TPP Program

#Question id: 14161

#Part-A Aptitude & General Biotechnology

Which of the following is not a description of dynamic programming algorithm?

TLS Online TPP Program

#Question id: 14164

#Part-B Specialized Branches in Biotechnology

Stoichiometric equations are used to represent growth of microorganisms provided a ' molecular formula' for the cells is available. The molecular formula for biomass is obtained by measuring the amounts of C, N, H, O and other elements in cells. For a particular bacterial strain, the molecular formula was determined to be C4.4H7.30 1.2N0.86 . These bacteria are grown under aerobic conditions with hexadecane (C16H34) as substrate. The reaction equation describing growth is:

                                  

  Assuming 100% conversion, what is the yield of cells from hexadecane in g g-l?   

TLS Online TPP Program

#Question id: 14165

#Part-B Specialized Branches in Biotechnology

Stoichiometric equations are used to represent growth of microorganisms provided a ' molecular formula' for the cells is available. The molecular formula for biomass is obtained by measuring the amounts of C, N, H, O and other elements in cells. For a particular bacterial strain, the molecular formula was determined to be C4.4H7.30 1.2N0.86 . These bacteria are grown under aerobic conditions with hexadecane (C16H34) as substrate. The reaction equation describing growth is:

                                    

Assuming 100% conversion, what is the yield of cells from oxygen in g g-l?